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Guys I didn't forget your request, just was collecting good questions to post.

So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.

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28 Aug 2015, 11:55

Bunuel wrote:

12. Is r=s?(1) -s<=r<=s (2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:A. s is either positive or zero, as -s<=s;B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5s=5, r=-5 --> -5<=-5<=5 |-5|>=5Both statements are true with these values. Hence insufficient.

Answer: E.

Can we eliminate B on the basis that it is just a reworded form of Statement A.

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28 Aug 2015, 12:29

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SauravPathak27 wrote:

Bunuel wrote:

12. Is r=s?(1) -s<=r<=s (2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:A. s is either positive or zero, as -s<=s;B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5s=5, r=-5 --> -5<=-5<=5 |-5|>=5Both statements are true with these values. Hence insufficient.

Answer: E.

Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.

(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT(2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT(1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT

Hopefully my freehand below makes it a little clearer.

If it helps.... Throw me a Kudos

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28 Aug 2015, 12:31

DropBear wrote:

SauravPathak27 wrote:

Bunuel wrote:

12. Is r=s?(1) -s<=r<=s (2) |r|>=s

This one is tough.

(1) -s<=r<=s, we can conclude two things from this statement:A. s is either positive or zero, as -s<=s;B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s.But we don't know whether r=s or not. Not sufficient.

(2) |r|>=s, clearly insufficient.

(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5s=5, r=-5 --> -5<=-5<=5 |-5|>=5Both statements are true with these values. Hence insufficient.

Answer: E.

Can we eliminate B on the basis that it is just a reworded form of Statement A.

My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.

(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT(2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT(1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT

Hopefully my freehand below makes it a little clearer.

If it helps.... Throw me a Kudos

Indeed it helps. Thanks for the extra effort though!
_________________

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Are 1 and to clearly insufficient because you can see at the first sight that x and y could be 0, leading to a clear No and that some positive values would immeadetly lead to a Yes. Can the case "won't be true" classify as a No to the question "is x^2+y^2>4a" or is it neutral?
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(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

Are 1 and to clearly insufficient because you can see at the first sight that x and y could be 0, leading to a clear No and that some positive values would immeadetly lead to a Yes. Can the case "won't be true" classify as a No to the question "is x^2+y^2>4a" or is it neutral?

Yes, you are correct. When you combine both the statements, you get \(x^2+y^2 = 5a\).

Consider the following 2 cases:

Case 1: When a = -1, 4a=-4 and 5a = -5. In this case you get a "NO"

Case 2: When a = 1, 4a=4 and 5a = 5. In this case you get a "YES"

Takeaway from this question: make sure to check the ranges of the variables!

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08 Sep 2015, 10:23

Bunuel wrote:

4. Are x and y both positive?(1) 2x-2y=1(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

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08 Sep 2015, 10:30

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reto wrote:

Bunuel wrote:

4. Are x and y both positive?(1) 2x-2y=1(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

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(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

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09 Sep 2015, 09:31

Bunuel wrote:

7. |x+2|=|y+2| what is the value of x+y?(1) xy<0(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2. >> Both x and y can be -2 leading to -4. So why this statement that either x or y is less than -2 and the other is more?
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09 Sep 2015, 09:40

reto wrote:

Bunuel wrote:

7. |x+2|=|y+2| what is the value of x+y?(1) xy<0(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2. >> Both x and y can be -2 leading to -4. So why this statement that either x or y is less than -2 and the other is more?

Because if x and y are both = -2, then you would end up with case A (x=y) and NOT case B.

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26 Sep 2015, 11:34

Bunuel wrote:

7. |x+2|=|y+2| what is the value of x+y?(1) xy<0(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

I have a doubt about this problema.I´m clear that x+y=-4, but at the end you can have as you said that x and y have differnts signs but also to have the same for example:

x= -2 and y = -2 then -2+(-2) = 4

So you can have x and y with differents signs or equals sings ans still be able to commit with the equation x+y = -4

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26 Sep 2015, 12:43

alfonsecar wrote:

Bunuel wrote:

7. |x+2|=|y+2| what is the value of x+y?(1) xy<0(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

I have a doubt about this problema.I´m clear that x+y=-4, but at the end you can have as you said that x and y have differnts signs but also to have the same for example:

x= -2 and y = -2 then -2+(-2) = 4

So you can have x and y with differents signs or equals sings ans still be able to commit with the equation x+y = -4

Your question is not clear. Where is it written that x and y can have same and different signs? Can you repost your question as it is difficult to understand what exactly are you trying to say.

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04 Oct 2015, 12:56

mrslee wrote:

Guys,

Is it ok plugging in in data sufficiency?

Yes, but you need to recognize where plugging in will be required and where algebraic equations will work. This realization can only come from practice. Try to solve some of the DS questions that you have solve using algebra by plugging in certain values and see which one gives you correct answer in average time frame.